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Unformatted text preview: 1 University of California, Berkeley Physics H7C Spring 2011 ( Yury Kolomensky ) SOLUTION TO PROBLEM SET 2 Special Relativity Maximum score: 200 points Due: February 4, 2011 Composed and formatted by P. Pebler, Yu. Kolomensky, and T.R. Beals 1. (30 points) Starting with the transformation for velocities and the definition of relativistic energy and momentum, prove that energy E and momentum pc of a given particle fol low Lorentz transformations, and form a relativistic 4 vector. 1. Solution Consider an inertial reference frame O , in which the particles velocity be u . In this frame, the particles energymomentum 4vector is pc = ( u mc 2 , u mu x c, u mu y c, u mu z c ) , where u := (1 u 2 /c 2 ) 1 / 2 . Now, let us consider a second in ertial reference frame O , moving with velocity v with respect to O . Without loss of generality, we can assume that v = v x , but we cannot assume that u  v that is, we cannot assume that u and v are colinear (for details on why we can assume v lies in the xdirection, see the solution to problem 10 from problem set 1). In the frame O , our particle has some new velocity u : u = v + u  + p 1 v 2 /c 2 u 1 + v u /c 2 = ( v + u x ) x + p 1 v 2 /c 2 ( u y y + u z z ) 1 + vu x /c 2 , and thus has a new energymomentum 4vector: p c = ( u mc 2 , u mu x c, u mu y c, u mu z c ) = u mc c, v + u x 1 + vu x c 2 , u y q 1 v 2 c 2 1 + vu x c 2 , u z q 1 v 2 c 2 1 + vu x c 2 , where u is as follows: u = " 1 u 2 c 2 # 1 / 2 = 1 ( v + u x ) 2 + (1 v 2 c 2 )( u 2 y + u 2 z ) (1 + vu x c 2 ) 2 c 2  1 / 2 = 1 + vu x c 2 c q c 2 v 2 u 2 + u 2 v 2 c 2 = 1 + vu x c 2 q 1 v 2 c 2 q 1 u 2 c 2 = u v 1 + vu x c 2 , where v = 1 / q 1 v 2 c 2 . Thus, we see that the particles new energymomentum 4vector is just the Lorentz transform of the old 4vector: p c = u v h mc 2 + mu x v i , v m ( v + u x ) c,mu y c,mu z c . Its also worth noting that q p p /c 2 = m , or, in other words, the invariant quantity preserved by Lorentz transformations of the energymomentum 4vector is the rest mass , m . 2. (20 points) A charged particle of charge e and momentum p is trav eling in a uniform magnetic field B p . Show that the trajectory is a circle of radius R , and that the radius and momentum are related by the expression p = eBR (in SI units). 2. Solution 2 To keep this problem easy, just do everything in the lab frame. All the usual laws of nonrelativistic mechanics apply, except that we use m in place of m . Thus, the force on a charge e moving with velocity v = p / ( m ) in a magnetic field B is just F = m a = e p m B ....
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This note was uploaded on 03/28/2012 for the course PHYSICS H7c taught by Professor Staff during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Staff
 Physics, Special Relativity

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